An equal volume of reducing agent is titrated separately with 1 M `KMnO_4` in acidic, neutral and alkaline media, the volumes of `KMnO_4` required are 20 mL in acid, 33.4 mL in neutral, and 100 mL in alkaline media. Find the oxidation state of Mn in each reduction product. Give balanced equation for all the three half reaction.Find the volume of `1 M K_2Cr_2O_7` consumed when the same volume of reducing agent is titrated in acidic medium.

Explanation

Let V mL of reducing agent be used for `KMnO_(4)` in different medium which act as oxidant
Acid medium, `Mn^(7+) + n_(2)e^(-) rarr Mn^(a+)`
`n_(1) = 7 - a `
Neutral medium, `Mn^(7+) + n_(2)e^(-) rarr Mn^(b+)`
`n_(2) = 7 - b `
Alkaline medium, `Mn^(7+) + n_(3)e^(-) rarr Mn^(c+)`
`n_(3) = 7 - c `
Meq. of reducing agent = Meq. of `KMnO_(4)` in acid medium = Meq. of `KMnO_(4)` in neutral
= Meq. of `KMnO_(4)` in alkaline = `1 xx n_(1) xx 20 = 1 xx n_(2) xx 33.3 = 1 xx n_(3) xx 100`
Since `n_(1), n_(2), n_(3)` are integers and `n_(1) < 7 `
`:. n_(1) = 5 , n_(2) = 3 and n_(3) = 1`
Therefore, different oxidation states of Mn are :
Acid media `Mn^(7+) + 5e^(-) rarr Mn^(a+) :. a = +2 `Neutral media `Mn^(7+) + 3e^(-) rarr Mn^(b+) :. b = +4`Alkaline media `Mn^(7+) + 1 e^(-) rarr Mn^(c+) :. c = +6 `Now same volume of reducing agent is treated with `K_(2)Cr_(2)O_(7)` and therefore,
Meq. of reducing agent = Meq. of `K_(2)Cr_(2)O_(7)`
`20 xx 5 = 1 xx 6 xx V`
`V = 100/6 = 16.66 mL`
`6e^(-) + Cr_(2)^(12+) rarr 2Cr^(3+) :. 1 M = 6 xx 1N`
It is important to note that the conditions are valid only when Mn in each medium exist as monoatomic, i.e., not as `Mn_(2)`